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Anna71 [15]
3 years ago
11

Which is hotter, Canopus or Vega? How much brighter?

Physics
1 answer:
marusya05 [52]3 years ago
4 0

Temperature determines the hotness or coldness. The surface temperature of Canopus is 7350 K. The surface temperature of Vega is 9602 K. Because the surface temperature of Vega is greater than Canopus, Vega is hotter.

Brightness is determined by the star's magnitude. The apparent magnitude of Canopus is (-0.72) and that of Vega is (0.03). we know that, lesser the magnitude, brighter is the star. Hence, Canopus is brighter.

Vega is hotter but Canopus is brighter. This is because, brightness depends on the star's distance and size of the star as well other than the factor temperature.

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The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies
mestny [16]

Complete question is;

The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies that EM radiation can have as the electromagnetic spectrum. Electromagnetic waves are used extensively in modern wireless technology. Many devices are built to emit and/or receive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are some examples followed by their frequencies of operation:

garage door openers: 40.0 MHz

standard cordless phones: 40.0 to 50.0 MHz

baby monitors: 49.0 MHz

FM radio stations: 88.0 to 108 MHz

cell phones: 800 to 900 MHz

Global Positioning System: 1227 to 1575 MHz

microwave ovens: 2450 MHz

wireless internet technology: 2.4 to 2.6 GHz

Which of the following statements correctly describe the various applications listed above? Check all that apply.

a.) All these technologies use radio waves, including low-frequency microwaves.

b.) All these technologies use radio waves, including high-frequency microwaves.

c.) All these technologies use a combination of infrared waves and high-frequency microwaves.

d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.

e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.

f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.

g.) All the technologies emit waves with a wavelength in the range of 0.01 to 10.0 km.

Answer:

B, D, E, F are the correct statements.

Explanation:

Looking at the options;

A) This is true because radio waves are electromagnetic radiation being used today in television, mobile phones, radios and other areas of communication technologies. And the examples given to us fall in the category of technologies that use radio waves.

B) microwaves usually have long wavelengths and low frequencies. However, sometimes they could have high frequencies usually more than radio waves. Thus, this option is correct.

C) This option is wrong because it's not all the listed technologies that use combination of infrared waves and high-frequency microwaves.

D) we are given the frequency of microwave ovens as 2450 MHz.

Converting to GHz gives; 2.45 GHz.

We are told that wireless internet technology has frequency between 2.4 to 2.6 GHz. Thus, microwave frequency falls in the same range as wireless internet technology and thus the statement is true.

E) we know that wavelength is inversely proportional to frequency. This means that the higher the frequency, the shorter the wavelength.

In the frequencies given to us, wireless internet technology have the highest frequency which means they have the shortest wavelength. The statement is true.

F) from the frequencies given to us, the smallest is garage door openers = 40.0 MHz = 40 × 10^(6) Hz while the biggest is 2.6 GHz = 2.6 × 10^(9) Hz

Formula for wavelength is;

Wavelength = speed of light/frequency

Speed of light = 3 × 10^(8) m/s

Thus;

Wavelength = (3 × 10^(8))/(40 × 10^(6))

Or wavelength = (3 × 10^(8))/(2.6 × 10^(9))

So,wavelength = 7.5 m or 0.12 m

This falls into the given range of 0.10 to 10.0 m.

Thus, the statement is true.

8 0
3 years ago
"a horn emits a frequency of 1000 HZ. A 14 m/s wind is blowing toward a listener. What is the frequency of the sound heard by th
Semenov [28]

Answer:

 f_L = 1000 Hz

Explanation:

given,

speed of wind, = 14 m/s

frequency of horn,f_s = 1000 Hz

speed of sound,V = 344 m/s

frequency heard by the listener

using Doppler effect

f_L = \dfrac{v+v_L}{v+v_s}f_s

f_L is the frequency of the sound heard by the listener

f_s is the frequency of sound emitted by the listener

V is the speed of sound

v_L is the speed of listener

v_s is the speed of source

now,

considering the frame of reference in which wind is at rest now, both listener and the source will be moving at 14 m/s

 f_L = \dfrac{v+14}{v+14}\times 1000

now on solving we will get

 f_L = 1000 Hz

hence, the frequency heard by the listener is equal to  1000 Hz

 

7 0
3 years ago
What is machine<br>and type of machine​
vitfil [10]
Printer could be a machine
3 0
4 years ago
Read 2 more answers
A spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. How fast is the radius of the balloon
Nookie1986 [14]

Answer:

<h2>0.245cm/min</h2>

Explanation:

The volume of the spherical balloon is expressed as V = 4/3πr³ where r is the radius of the spherical balloon. If the spherical balloon is inflated with gas at the rate of 500 cubic centimetres per minute then dV/dt = 500cm³.

Using chain rule to express dV/dt;

dV/dt = dV/dr*dr/dt

dr/dt is the rate at which the radius of the gallon is increasing.

From the formula, dV/dr = 3(4/3πr^3-1))

dV/dr = 4πr²

dV/dt = 4πr² *dr/dt

500 =  4πr² *dr/dt

If radius r = 40;

500 = 4π(40)² *dr/dt

500 = 6400π*dr/dt

dr/dt = 500/6400π

dr/dt = 5/64π

dr/dt  = 0.245cm/min

Hence, the radius of the balloon is increasing at the rate of 0.245cm/min

8 0
3 years ago
Heat from burning fuel warms the walls of the firebox section of the furnace in
nydimaria [60]

Heat from burning fuel warms the walls of the firebox section of the furnace in

A. a hot-water heating system.

B. a hot-air heating system.

C. a compressor compartment.

D. an evaporation system.

7 0
3 years ago
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